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i)x^2-2x+A is positive for all x ii)Ax^2+1 is positive for all x reason given < Statement (1) by itself is sufficient. x^2-2x+A = (x-1)^2+(A-1) . For this expression to be always positive (A-1) has to be more than 0.so A has to be more than 1>A can be any -ive int till (x-1)^2 is greater than (A-1)... ex x=3... 4+A-1>0.. or A>-3 ans shud be E...pl confirm i think where we have gone wrong is taking + in(x-1)^2+(A-1) as *....
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i)x^2-2x+A is positive for all x ii)Ax^2+1 is positive for all x reason given < Statement (1) by itself is sufficient. x^2-2x+A = (x-1)^2+(A-1) . For this expression to be always positive (A-1) has to be more than 0.so A has to be more than 1>A can be any -ive int till (x-1)^2 is greater than (A-1)... ex x=3... 4+A-1>0.. or A>-3 ans shud be E...pl confirm i think where we have gone wrong is taking + in(x-1)^2+(A-1) as *....

Step by step:

Question: is \(A>0\)?

(1) \(x^2-2x+A\) is positive for all \(x\):

\(f(x)=x^2-2x+A\) is a function of of upward parabola. We are told that it's positive for all \(x\) --> \(f(x)=x^2-2x+A>0\), which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation \(x^2-2x+A=0\) has no real roots.

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hi ..... i do understand that as an eq of parabola c would be y-intercept and parabola being limited to i and ii quad , c would be always +ive...... but why cant we prove it taking it as a normal eq (as it wud seem to any layperson)... is there any way to prove it
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